Question

Find the modulus and argument of the complex number:
$\frac{1+i}{1-i}$

Answer 1

Let $z=\frac{1+i}{1-i}$

Rationalizing the same,

$z=\frac{1+i}{1-i}\ast \frac{1+i}{1+i}$

$z=\frac{(1+i)(1+i)}{(1-i)(1+i)}$

$Using(a-b)(a+b)=a^2-b^2$

$=\frac{{(1+i)}^2}{1^2-i^2}$

$Using{(a+b)}^{2=}{a}^2+b^2+2ab$

$=\frac{1^2+i^2+2i}{1^2-i^2}$

$Putting{i}^2=-1$

$=\frac{1^2+(-1)+2i}{1-(-1)}$

$=\frac{2i}{2}$

$=i$

$=0+i$

$Hencez=(0+i)$

To calculate modulus of z,

$z=(0+i)$

$Complexnumberzisoftheformx+iy$

$Hencex=0andy=1$

Modulus of $z=\sqrt{x^2+y^2}$

$z=\sqrt{0^2+1^2}$

$z=\sqrt{0+1}$

$z=\sqrt{1}$

To find the argument,

$0+i=rcos\theta +irsin\theta$

Comparing real part,

$0=rcos\theta$

Put r=1

$0=1\ast cos\theta$

$0=cos\theta$

$cos\theta =0$

Comparing imaginary part,

$1=rsin\theta$

Put r=1

$1=1\ast sin\theta$

$1=sin\theta$

$sin\theta =1$

$Hencecos\theta =0andsin\theta =1$